Question about car engines

[agp423]

I was talking to a friend yesterday who has a 1998 Mustang with a 3.8L V6 engine, which I thought would have a higher hp/torque output than my 3.6L V6. But it turns out that his car only has 150hp whereas mine has 255hp. How can a 3.8L V6 be less powerful than 3.6L V6?

[dave_from_here]

Re: Question about car engines

Many factors determine an engine's power. Size is one of them but so is cam lift & profile, computer programing, exhaust restriction, is the car turbo charged... etc..

H.P. comes from a mathematical formula : Torque*rpm /5250

Where Torque is the amount of Work that an engine can do
RPM is the speed at witch the engine works
5250 is a constant used for the formula (I don't remember why)


so if engine A revs faster than engine B it 's possible that it will make more horsepower even if smaller

[dietoremain]

Re: Question about car engines

Yeah size helps but it doesnt always matter..
Heck my truck puts down 250HP and 525TQ but diesels are a whole 'nother world... haha. For example the 6.0 ford puts out 325HP and like 550TQ but its much smaller. Turbo chargers help. Its also in the engines design, compression ratio, etc.

[paudashlake]

Re: Question about car engines

My minivan has a 3.6l v6 with 210hp, but it is a lot more to haul around than a little car.

[IcedTea]

Re: Question about car engines

Put very simply, your LY7 is DOHC and is much more efficent in putting out power than his OHV V6. Of course, there are more factors such as compression ratio etc.

Displacement isn't everything (Coming from a guy who drives a 6 liter V8)

[the other pc]

Re: Question about car engines

...5250 is a constant used for the formula (I don't remember why)...

5250 is a conversion factor used because there’s a difference in units in the equation.

Just as you you’d divide the number of pints in a glass by eight to get the number of gallons or multiply a length in inches by 2.54 to get that length in centimeters, you throw that 5250 (approximately) in when you’re using the units that are commonly found in American gearhead literature.

You see that equation in hotrod and car books all the time but you rarely see it mentioned that it only works for torque measured in foot-pounds. And it only works for power measured in horsepower.

If you’re measuring torque in any other units, Nm, pound-inches, dyne centimeters or whatever, 5250 doesn’t work.

If you’re looking for power in any other units, Watts, kilowatts, Btu/min, refrigeration tons or whatever, 5250 doesn’t work.



Warning! Science content. Read at your own risk.


Power is defined as the ability to do work over time.

P = W/time

Work is defined as a force applied over a distance.

W = F*dist

If you lift a 1 lb weight up a foot, you’ve done 1 ftlb of work.

If you do it in one minute, you’ve applied 1 ftlb/min of power over that minute.

Torque is twisting force. It’s defined as a force applied at a radius.

T = F*radius

Torque only tells you how much force you’re applying. To know how much power you can exert and how much work you can accomplish, you need to know how fast you can move while applying the force.

If we want to know how much power we’re generating we need to know how much work we’re accomplishing and how long it’s taking.

since
P = W/time
And
W = F*dist

P = W/time = F*dist/time

To find the distance traveled and time:

We know speed:
speed = rev/min

We know the torque
T = F*radius

We know that the distance we’ve traveled in one revolution is the circumference of the circle we’re applying the torque to.

dist/rev = 2pi* radius

therefore:

dist = 2pi*radius*rev


since we know that P = F*dist/time
then
P = F*dist/time = (F*2pi*radius*rev)/time

Rearranging that a bit:

P = (F*radius)*(rev/time)*2pi

Since we know that T = F*radius and that speed in rpm = rev/time, we know that:

Drumroll, please….. P = T*rpm*2pi

And this expresses the power in ftlb/min for torques expresses in ftlb.

But since we want power expressed in horsepower, we need a conversion factor. Conversion factors are standardized by agreement between standardizing organizations (and the exciting scientific folks they employ) around the world.

When you look up the factor you’ll find that 1 hp = 33000 ftlb/min

P{hp} = P{ftlb/min}/33000

So

P{hp} = T*rpm*2pi/33000

And since 2pi/33000 = 1/ 5252.11

P{hp} = T{ftlb}*rpm/ 5252.11


dot, dot, dot.



pc.

[dietoremain]

Re: Question about car engines

Paul you just fried my brain.

[ClearlyCoated]

Re: Question about car engines

Paul you just fried my brain.

Yeah, Paul. I just sprained a lobe. Thanks a heap.

[Markus Kleis]

Re: Question about car engines

I was talking to a friend yesterday who has a 1998 Mustang with a 3.8L V6 engine, which I thought would have a higher hp/torque output than my 3.6L V6. But it turns out that his car only has 150hp whereas mine has 255hp. How can a 3.8L V6 be less powerful than 3.6L V6?

The biggest reason is likely the time produced (ie when it was made) and cost of the vehicle.

You're comparing a very old engine to a fairly new one, right?

There are early V8 engines that put out less than half of what many new 4 cylinder engines put out today, for example

[Andy M.]

Re: Question about car engines

I have to agree with Mark on this one. New technologies have brought about more HP from the smaller newer engines. Although the 3.8 litre V6 has pretty much stood still and is a bulletproof engine. They absolutely run forever when maintained with nothing but regular maintenance. Last I checked the GM 3.8 litre V-6 is rated at 200 hp. This was the same motor I had in a 1991 Buick Regal LS and still the same in my 2001 Impala LS, still 200 HP. For example look at the Honda Civic SI, with a whopping 197 HP out of a 4 banger. Aluminum components, lighter cars make them quicker. That all being said I am a car guy and I still say, "there is no replacement for displacement". Right now I have my Impala SS with a 5.3 litre V-8 or 327, for us older car guys boasting 303 HP. I'll stick with my SS as my top end is absolutely unbelievable!!!!

Bring on the Honda's!!! LOL

Andy

[roushstage2]

Re: Question about car engines

All of that and...Ford has been lacking in the giving of the HP department. My one gripe with 'em. The new 5.0 however, is a step in the right direction, and a big step indeed. If you are going to make a performance V8, make a performance V8.

[ColonelCash]

Re: Question about car engines

All of that and...Ford has been lacking in the giving of the HP department. My one gripe with 'em. The new 5.0 however, is a step in the right direction, and a big step indeed. If you are going to make a performance V8, make a performance V8.

My biggest gripe with Ford is for making cars for teenagers. Granted, the Cobra / Saleen / Roush lines are worthy of anyones money, but the 4 cyl bagrain mustangs annoy me to no end....maybe its the cheap plastic rims, I don't know.

[Markus Kleis]

Re: Question about car engines

My biggest gripe with Ford is for making cars for teenagers. Granted, the Cobra / Saleen / Roush lines are worthy of anyones money, but the 4 cyl bagrain mustangs annoy me to no end....maybe its the cheap plastic rims, I don't know.

Ford hasn't offered a 4 cylinder Mustang in almost 20 years

Back when they did they weren't that underpowered, either. In fact, the SVO Mustang (2.3-liter inline 4 with turbo) was a performance model that at one point was the highest output Mustang offered

[roushstage2]

Re: Question about car engines

Ford hasn't offered a 4 cylinder Mustang in almost 20 years

Back when they did they weren't that underpowered, either. In fact, the SVO Mustang (2.3-liter inline 4 with turbo) was a performance model that at one point was the highest output Mustang offered

The SVO was fine, but the regular ones had under 100RWHP, lol. Someone dynoed one at a dyno day and I think it was somewhere around 93RWHP. They have been long gone though.

[dave_from_here]

Re: Question about car engines

5250 is a conversion factor used because there’s a difference in units in the equation.

Just as you you’d divide the number of pints in a glass by eight to get the number of gallons or multiply a length in inches by 2.54 to get that length in centimeters, you throw that 5250 (approximately) in when you’re using the units that are commonly found in American gearhead literature.

You see that equation in hotrod and car books all the time but you rarely see it mentioned that it only works for torque measured in foot-pounds. And it only works for power measured in horsepower.

If you’re measuring torque in any other units, Nm, pound-inches, dyne centimeters or whatever, 5250 doesn’t work.

If you’re looking for power in any other units, Watts, kilowatts, Btu/min, refrigeration tons or whatever, 5250 doesn’t work.



Warning! Science content. Read at your own risk.


Power is defined as the ability to do work over time.

P = W/time

Work is defined as a force applied over a distance.

W = F*dist

If you lift a 1 lb weight up a foot, you’ve done 1 ftlb of work.

If you do it in one minute, you’ve applied 1 ftlb/min of power over that minute.

Torque is twisting force. It’s defined as a force applied at a radius.

T = F*radius

Torque only tells you how much force you’re applying. To know how much power you can exert and how much work you can accomplish, you need to know how fast you can move while applying the force.

If we want to know how much power we’re generating we need to know how much work we’re accomplishing and how long it’s taking.

since
P = W/time
And
W = F*dist

P = W/time = F*dist/time

To find the distance traveled and time:

We know speed:
speed = rev/min

We know the torque
T = F*radius

We know that the distance we’ve traveled in one revolution is the circumference of the circle we’re applying the torque to.

dist/rev = 2pi* radius

therefore:

dist = 2pi*radius*rev


since we know that P = F*dist/time
then
P = F*dist/time = (F*2pi*radius*rev)/time

Rearranging that a bit:

P = (F*radius)*(rev/time)*2pi

Since we know that T = F*radius and that speed in rpm = rev/time, we know that:

Drumroll, please….. P = T*rpm*2pi

And this expresses the power in ftlb/min for torques expresses in ftlb.

But since we want power expressed in horsepower, we need a conversion factor. Conversion factors are standardized by agreement between standardizing organizations (and the exciting scientific folks they employ) around the world.

When you look up the factor you’ll find that 1 hp = 33000 ftlb/min

P{hp} = P{ftlb/min}/33000

So

P{hp} = T*rpm*2pi/33000

And since 2pi/33000 = 1/ 5252.11

P{hp} = T{ftlb}*rpm/ 5252.11


dot, dot, dot.



pc.

Great post